Math 181 Miniproject 4: Linear Approximation and Calculus.md
---
Math 181 Miniproject 4: Linear Approximation and Calculus
===

**Overview:** In this miniproject you will put the idea of the *local linearization* of a function to build linear approximations to complex functions and then make *interpolations* and *extrapolations* using them.

**Prerequisites:** Sections 1.8 in *Active Calculus*, which focuses on this topic. **Completion of Miniprojects 1 and 2 is recommended before doing this miniproject**.

---

:::info
1\. A potato is placed in an oven, and the potato's temperature $F$ (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. The time $t$ is measured in minutes.
| $t$ | 0 | 15 | 30 | 45 | 60 | 75 | 90 |
|----- |---- |------- |----- |----- |------- |------- |------- |
| $F$ | 70 | 180.5 | 251 | 296 | 324.5 | 342.8 | 354.5 |

(a) Use a central difference to estimate $F'(75)$. Use this estimate as needed in subsequent questions in this problem.
:::

(a)
$F'(75)=\frac{60-90}{60-90}$

$F'(75)=\frac{324.5-354.5}{30}$

$F'(75)=1°F/min$
:::info
(b) Find the local linearization $y = L(t)$ to the function $y = F(t)$ at the point where $a = 75$.
:::

(b)
$L(x)=F(75)+F'(75)(x-75)$
$L(x)=342.8+1(x-75)$

:::info
(c\) Determine an estimate for $F(72)$ by employing the local linearization. Terminology: This estimate is called an *interpolation* because we are estimating a value that lies within a data set, between two known data points.
:::

(c\)
$L(x)=F(75)+F'(75)(x-75)$
$=342.8+1(x-75)$
$L(72)=342.8+1(72-75)$
$L(72)=342.8+1(-3)$
$L(72)=342.8-3$
$L(72)=339.8°F$

:::info
(d) Do you think your estimate in (c) is too large, too small, or exactly right? Why?
:::

(d)

Based from the graph the estimate is exaclty right. Because the value is touching the tangent line and not over or under estimating it.
:::info
(e) Use your local linearization to estimate $F(100)$. Terminology: This estimate is called an *extrapolation* because we are estimating a value that lies outside the range of values of a data set.
:::

(e)
$L(x)=F(75)+F'(75)(x-75)$
$=342.8+1(x-75)$
$L(100)=342.8+1(100-75)$
$L(100)=342.8+1(25)$
$L(100)=342.8+25$
$L(100)=367.8°F$

:::info
(f) Do you think your estimate in (e) is too large, too small, or exactly right? Why?
:::

(f)

The estimate is a bit large. Because based from the graph the function is gong concave down. Therefore, the value should be less From what I estimated because it is under-estimating it.

:::info
(g) Plot both $F$ and $L$ and comment on how or when the line $L(t)$ is a good approximation of $F(t)$.
:::

(g)
It is a good approximation when the tangent like is touching the function. However, when it is concave down or up approximation can vary because it is either under-estimating or over-estimating.

---

To submit this assignment click on the Publish button . Then copy the url of the final document and submit it in Canvas.